Question: Suppose we have a vector field $f(x, y) = \left( 2y, 3x \right)$ and a curve $C$ that is parameterized by $\alpha(t) = (\cos(t), \sin(t))$ for $-\pi < t < \pi$. What is the line integral of $f$ along $C$ ? $ \int_C f \cdot d\alpha = $
Explanation: Given a vector field $f$, a parameterization $\alpha$, and bounds $t_0$ and $t_1$, we can calculate the line integral as follows: $ \int_C f \cdot d\alpha = \int_{t_1}^{t_2} f(\alpha(t)) \cdot \alpha'(t) \, dt$ Here, $f(x, y) = \left( 2y, 3x \right)$ and $\alpha(t) = (\cos(t), \sin(t))$. $\begin{aligned} &f(\alpha(t)) = (2\sin(t), 3\cos(t)) \\ \\ &\alpha'(t) = (-\sin(t), \cos(t)) \end{aligned}$ Now we can rewrite our line integral as a single-variable integral. $ \int_C f \cdot d\alpha = \int_{-\pi}^{\pi} (2\sin(t), 3\cos(t)) \cdot (-\sin(t), \cos(t)) \, dt$ Let's solve the integral. We'll use the Pythagorean identity that $\cos^2(\theta) + \sin^2(\theta) = 1$ and an identity that $\sin^2(\theta) = \dfrac{1 - \cos(2\theta)}{2}$. $\begin{aligned} &\int_{-\pi}^{\pi} (2\sin(t), 3\cos(t)) \cdot (-\sin(t), \cos(t)) \, dt \\ \\ &= \int_{-\pi}^{\pi} -2\sin^2(t) + 3\cos^2(t) \, dt \\ \\ &= \int_{-\pi}^{\pi} -2\sin^2(t) + (3 - 3\sin^2(t)) \, dt \\ \\ &= \int_{-\pi}^{\pi} -5\sin^2(t) + 3 \, dt \\ \\ &= \int_{-\pi}^{\pi} -5 \left( \dfrac{1 - \cos(2t)}{2} \right) + 3 \, dt \\ \\ &= \left[ -5 \left( \dfrac{t}{2} - \dfrac{\sin(2t)}{4} \right) + 3t \right]_{-\pi}^{\pi} \\ \\ &= \left[ \dfrac{t}{2} - \dfrac{5\sin(2t)}{4} \right]_{-\pi}^{\pi} \\ \\ &= \left( \dfrac{\pi}{2} - \dfrac{5 \sin(2\pi)}{4} \right) - \left( \dfrac{-\pi}{2} - \dfrac{5\sin(-2\pi)}{4} \right) \\ \\ &= \pi \end{aligned}$ In conclusion, the line integral $ \int_C f \cdot d\alpha = \pi$.